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anonymous

  • 4 years ago

Evaluate the determinants to verify the equation

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  1. anonymous
    • 4 years ago
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  2. anonymous
    • 4 years ago
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    I have tried solving this and have been unsuccessful. Can anyone help me out?

  3. JamesJ
    • 4 years ago
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    is this your problem child of a problem?

  4. anonymous
    • 4 years ago
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    lol my problem

  5. anonymous
    • 4 years ago
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    ummm the equation after the equal sign. Is that the determinant?

  6. anonymous
    • 4 years ago
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    Since i know what the determinant equals but it doesnt equal that

  7. JamesJ
    • 4 years ago
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    it's asking you to evaluate the determinant and then show it does equal the expression on the RHS (right and side).

  8. anonymous
    • 4 years ago
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    well it doesnt LOL

  9. JamesJ
    • 4 years ago
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    sure? I find the determinant to be \[ bc^3 + ca^3 + ab^3 - (ba^3 + cb^3 + ac^3) \]

  10. anonymous
    • 4 years ago
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    ya i got that too

  11. JamesJ
    • 4 years ago
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    while the RHS, \[ RHS = (-a^2b + a^2c + ab^2 - ac^2 - b^2c + bc^2)(a+b+c) \] \[= ab^3 + bc^3 + ca^3 -(ba^3 - ac^3 - cb^3) \] which is the same.

  12. JamesJ
    • 4 years ago
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    in the brackets, should be + +

  13. anonymous
    • 4 years ago
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    LOL I dont know where I went wrong but when I foiled I Landed up with too many cs

  14. JamesJ
    • 4 years ago
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    cs?

  15. anonymous
    • 4 years ago
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    like the variable c

  16. JamesJ
    • 4 years ago
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    This problem is training you. Time to move beyond FOIL and multiply things out and keep track of them. Work this again and make sure you can replicate the result.

  17. anonymous
    • 4 years ago
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    like what do u mean by moving beyond foil?

  18. JamesJ
    • 4 years ago
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    I mean you shouldn't need the rubric anymore. And what's more, you are seeing things with more than two terms in each bracket now. You should teach yourself to do that without using foil, a la the following (x + y + z)(a + b + c) = xa + xb + xc + ya + yb + yc + za + zb + zc

  19. anonymous
    • 4 years ago
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    ya but like there it was (a-b)(b-c)(c-a)

  20. anonymous
    • 4 years ago
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    so first i did (a-b)(b-c) and then i multiplied my answer by (c-a) Is that correct?

  21. JamesJ
    • 4 years ago
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    exactly so, that's equal to (ab - ac -b^2 + bc)(c-a) = abc - ac^2 -b^2c + bc^2 -a^2b+a^2c + ab^2 - abc = ab^2 + bc^2 + ca^2 - (ac^2 + ba^2 + ac^2)

  22. JamesJ
    • 4 years ago
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    etc., etc. Lots of ways to skin this cat. But having the discipline to work through it is important.

  23. anonymous
    • 4 years ago
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    okkkkk i just caught my mistake LOL

  24. JamesJ
    • 4 years ago
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    ok, I'm going to move on.

  25. anonymous
    • 4 years ago
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    You make one silly mistake and it ruins everything. Thanks James for ur help

  26. JamesJ
    • 4 years ago
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    Yes. Here's a hint for this problem. Clearly the problem is symmetric in a,b,c. Which means if we swapped the order of them it shouldn't upset the result too much. That being the case, at every step of the way, we should expect a certain symmetry in the terms. So if there's a term such as ab^2 it make sense that we should also expect a term such as bc^2 and ca^2 This is a way to at least sense check your answer at each step.

  27. anonymous
    • 4 years ago
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    Alrighty Thanks for ur help :DDDDDDDD

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